Past exam of the martingales and financial mathematics course of the University of Oxford / 2010 / A / 1

A measurable space is a set $E$ together with a sigma-algebra $\mathcal{E}$ of its subsets: it contains $E$ and is closed under complements relative to $E$ and countable unions. A measure is a map $\mu :\mathcal{E}\to [0,\infty ]$ with $\mu (\varnothing )=0$ and countable additivity on disjoint measurable sets:A finite measure additionally satisfies $\mu (E)<\infty$.

For $\epsilon >0$, define the measurable decreasing tail eventsIf $f_n\to f$ almost everywhere, every point outside its null exceptional set eventually leaves each such tail event, so $\mu ({\bigcap }_n{A}_n(\epsilon ))=0$. Continuity from above of a measure applies because $\mu (E)<\infty$, giving $\mu (A_n(\epsilon ))\to 0$.

Conversely, assume these measures tend to zero. For each positive integer $k$, the intersection ${\bigcap }_n{A}_n(1/k)$ has measure zero. Outside the union of these countably many null sets, every $k$ has an index after which $|f_m-f|\le 1/k$ for all later $m$. Hence $f_m\to f$ there. This proves the desired equivalence. The argument uses all positive thresholds; replacing the intersection of tail events by a single strict limsup inequality at the same threshold would mishandle equality cases.

The triangle inequality gives the event inclusionCountable subadditivity of a measure bounds its measure by the sum of the two terms on the right. Both tend to zero as $m,n\to \infty$ by convergence in measure. Thus convergence in measure implies the Cauchy property in measure.

The Cauchy assumption lets us choose $n_k$ recursively, increasing strictly, so that $\mu (|f_n-f_m|>2^{-k})<2^{-k}$ for all $m,n\ge n_k$. In particular the measurable eventshave summable measures. The measure of ${\bigcup }_{k\ge K}{E}_k$ is at most ${\sum }_{k\ge K}{2}^{-k}\to 0$, so $\mathrm{limsup}{E}_k$ is a null set. Outside it, the successive differences are eventually bounded by a summable geometric series. Hence $f_{n_k}$ is pointwise Cauchy and has a finite real limit.

Define $f$ to be that limit wherever the sequence converges, and zero elsewhere. The convergence set is measurable by the countable Cauchy criterion, and the limit is measurable as a limit of measurable functions on that set. Thus there is no need to assume the measure is complete. Part (a)(ii), or the corresponding finite-measure implication, gives $f_{n_k}\to f$ in measure.

For the full sequence useChoose a sufficiently late subsequence index so the second term is small; the Cauchy property then bounds the first uniformly for all large $n$. Therefore $f_n\to f$ in measure. This proves completeness of convergence in measure on a finite measure space.